prove that the absolute valje of z1+z2<=absolute value of z1+absolute value of z2


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Question Number 139778 by chiamaka last updated on 01/May/21

$p r o v e t h a t t h e a b s o l u t e v a l j e o f z 1 + z 2 <= a b s o l u t e v a l u e o f z 1 + a b s o l u t e v a l u e o f z 2$
	Answered by mr W last updated on 01/May/21

	Commented by mr W last updated on 01/May/21

	$O A = z_{1}$ $O B = z_{2}$ $O C = z_{1} + z_{2}$ $∣ A C ∣=∣ O B ∣=∣ z_{2} ∣$ $∣ O C ∣⩽∣ O A ∣ + ∣ A C ∣$ $\Rightarrow∣ z_{1} + z_{2} ∣⩽∣ z_{1} ∣ + ∣ z_{2} ∣$
	Answered by mr W last updated on 01/May/21

	$z_{1} = a + b i$ $∣ z_{1} ∣= \sqrt{a^{2} + b^{2}}$ $z_{2} = p + q i$ $∣ z_{2} ∣= \sqrt{p^{2} + q^{2}}$ $z_{1} + z_{2} = (a + p) + (b + q) i$ $∣ z_{1} + z_{2} ∣= \sqrt{{(a + p)}^{2} + {(b + q)}^{2}}$ ${M i n k o w s k i}^{'} s i n e q u a l i t y :$ $∣ \sqrt{{(a + p)}^{2} + {(b + q)}^{2}} ⩽ \sqrt{a^{2} + b^{2}} + \sqrt{p^{2} + q^{2}}$ $i . e .$ $∣ z_{1} + z_{2} ∣⩽∣ z_{1} ∣ + ∣ z_{2} ∣$
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